Percentage Calculator
Calculate percentage from obtained marks and total marks.
This chapter belongs to a previous NCERT/CBSE curriculum and is not part of the current 2026-27 chapter list. Removed in 2022-23.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter States of Matter. This page offers a step-by-step solution to the specific question from Exercise 1, Question 12:
Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
(R = 0.083 bar dm3 K–1 mol–1).
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar.
(R = 0.083 bar dm3 K–1 mol–1).
Given,
Amount of the gas, n = 4.0 mol
Volume of the gas, V = 5 dm3
Pressure of the gas, p = 3.32 bar
R = 0.083 bar dm3 K–1 mol–1
From the ideal gas equation, we get
pV = nRT
⇒ T = pV / nR
= (3.32x5) / (4x0.083)
= 50K
Hence, the required temperature is 50 K.
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 12: Calculate the temperature of 4.0 mol of a gas occupying 5 dm3 at 3.32 bar. (R = 0.083 bar dm3 K&n....
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Comments
Nice and easy solution , thanks
Tq a lot . It helped me a lot . Upload question from subject chemistry of 2nd puc. Waiting.............
Its osm tqsm
Please can you give me further answers???
pressure of gas filled in a closed cylinder 273k is 4 atm . calulate the temp of the gas when its pressure is 5 atm pleas send me the ans as soon as posible